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I have to prove the following theorem:

Theorem: Let $f:(M,d)\to (N,\rho)$ be one-to-one and onto. Then the following are equivalent:

i) $f$ is a homeomorphism.

ii) $x_n\stackrel{d}{\to}x \Leftrightarrow f(x_n)\stackrel{\rho}{\to}f(x).$

iii) $G$ is open in $M \Leftrightarrow f(G)$ is open in $N$.

iv) E is closed in $M \Leftrightarrow f(E)$ is closed in $N$.

v) $\hat{d}(x,y) = \rho(f(x),f(y))$ defines a metric on $M$ equivalent to $d$.

What I've done so far:

i) $\Rightarrow$ iii) Because $f$ is a homeomorphism, we have that $f$ and $f^{-1}$ are continuous meaning that a set $G$ is open in $M$ iff $f(G)$ is open in $N$.

iii) $\Rightarrow$ iv) Saying that $G$ is open in $M$ iff $f(G)$ is open in $N$, is equivalent to saying that $M\backslash G$ is closed in $M$ iff $N\backslash F(G)$ is closed in $N$.

ii) $\Rightarrow$ v) We have that $d(x_n,x)\to 0$ iff $\rho(f(x_n),f(x))\to 0$ and hence $d(x,y)$ and $\hat{d}(x,y)$ are equivalent metrics on $M$.

v) $\Rightarrow$ i) If $d$ and $\rho$ are equivalent on $M$ then there exists a homeomorphism from $(M,d)$ to $(N,\rho)$.

I can't figure out how I should prove iv) $\Rightarrow$ ii). I've tried to use a different order in proving the equivalent statements but I can't think of one.

Question: Is my approach correct thus far? How should I prove the last part?

Thanks!

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It is sufficient to show that $f$ is continuous. As both spaces are metric-spaces so they are first countable and so $f$ will be continuous $\iff$ (ii) holds.

Let, $U$ be an open set in $(N,\rho)$. Then $F=N-U$ is a closed set in $(N,\rho)$.

By, (iv) $f^{-1}(F)$ is closed in $(M,d)$. Since $f$ is a bijection so,

$f^{-1}(F)=f^{-1}(N-U)=M-f^{-1}(U)$ is closed in $(M,d)$.

Consequently, $f^{-1}(U)$ is open in $(M,d)$, so $f$ is continuous. Hence the theorem.

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